• 最大子数组和问题的解


    寻找数组A的和最大的非空连续子数组。例如:A[] = {1, -2, 3, 10, -4, 7, 2, -5}的最大子数组为3, 10, -4, 7, 2,其最大和为18。数组元素都为负时,返回最大负值。

    解法一:暴力枚举,O(N^2)

     1 //left,最大子数组左端
     2 //right,最大子数组右端
     3 //n数组arr元素个数
     4 const int IntMin = (signed int)0x80000000;
     5 int MaxSum(int *arr, int &left, int &right, int n)
     6 {
     7     if (!arr || n <= 0) {
     8         left = -1;
     9         right = -1;
    10         return IntMin;
    11     }
    12 
    13     int maxSum = IntMin;
    14     for (int i = 0; i < n; ++i) {
    15         int sum = 0;
    16         for (int j = i; j < n; ++j) {
    17             sum += arr[j];
    18             if (sum > maxSum) {
    19                 left = i;
    20                 right = j;
    21                 maxSum = sum;
    22             }
    23         }
    24     }
    25     return maxSum;
    26 }

    解法二:分治,O(NlogN)

    寻找数组A[low…high]的最大子数组时,将数组划分为两个规模尽可能相等的子数组 A[low…mid]和A[mid+1…high]。

    A[low…high]的任何连续子数组A[i…j]所处的位置必然是三种情况之一:

    (1)完全位于子数组A[low..mid]中,因此low<=i<=j<=mid; 

    (2)完全位于子数组A[mid + 1..high]中,因此mid<=i<=j<=high;

    (3)跨越了中点,因此low<=i<=mid<j<=high;

    A[low…high]的一个最大子数组必然是完全位于A[low…mid]中、完全位于A[mid+1…high]中或者跨越中点的所有子数组中和最大者。

     1 //left和right是返回值,未初始化,不能直接使用
     2 const int IntMin = (signed int)0x80000000;
     3 int MaxSumCross(int *arr, int &left, int mid, int &right, int n)
     4 {
     5     if (arr == NULL) {
     6         left = -1;
     7         right = -1;
     8         return IntMin;
     9     }
    10 
    11     int lmaxSum = IntMin;
    12     int sum = 0;
    13     for (int i = mid; i >= 0; --i) {
    14         sum += arr[i];
    15         if (sum > lmaxSum) {
    16             left = i;
    17             lmaxSum = sum;
    18         }
    19     }
    20 
    21     int rmaxSum = IntMin;
    22     sum = 0;
    23     for (int i = mid + 1; i < n; ++i) {
    24         sum += arr[i];
    25         if (sum > rmaxSum) {
    26             right = i;
    27             rmaxSum = sum;
    28         }
    29     }
    30     return lmaxSum + rmaxSum;
    31 }
    32 
    33 int MaxSum(int *arr, int &left, int &right, int n)
    34 {
    35     if (!arr || left > right || n <= 0) {
    36         left = -1;
    37         right = -1;
    38         return IntMin;
    39     }
    40     if (left == right) 
    41         return arr[left];
    42 
    43     //lleft,lright左子数组的左端,右端
    44     //rleft,rright右子数组的左端,右端
    45     //mleft,mright中间子数组的左端,右端
    46     int mid = left + (right - left) / 2;
    47     int lleft = left, lright = mid;
    48     int rleft = mid + 1, rright = right;
    49     int mleft, mright;
    50     int leftSum = MaxSum(arr, lleft, lright, n);
    51     int midSum = MaxSumCross(arr, mleft, mid, mright, n);
    52     int rightSum = MaxSum(arr, rleft, rright, n);
    53 
    54     if (leftSum >= rightSum && leftSum >= midSum) {
    55         left = lleft;
    56         right = lright;
    57         return leftSum;
    58     } else if (rightSum >= leftSum && rightSum >= midSum) {
    59         left = rleft;
    60         right = rright;
    61         return rightSum;
    62     } else if (midSum >= leftSum && midSum >= rightSum) {
    63         left = mleft;
    64         right = mright;
    65         return midSum;
    66     }
    67 }

    解法三:动态规划,O(N)

     1 int MaxSum(int *arr, int &left, int &right, int n)
     2 {
     3     if (!arr || n <= 0) {
     4         left = -1;
     5         right = -1;
     6         return IntMin;
     7     }
     8 
     9     int nAll = arr[n - 1];
    10     int nStart = arr[n - 1];
    11     left = 0;
    12     right = n - 1;
    13     for (int i = n - 2; i >= 0; --i) {
    14         if (nStart < 0) {
    15             right = i;
    16             nStart = 0;
    17         }
    18         nStart += arr[i];
    19         if (nStart > nAll) {
    20             left = i;
    21             nAll = nStart;
    22         }
    23     }
    24     return nAll;
    25 }
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  • 原文地址:https://www.cnblogs.com/mengwang024/p/4627263.html
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